Question: $f(x, y) = (xy - 1, y(\cos(x) + x))$ $\text{curl}(f) = $
The formula for curl in two dimensions is $\text{curl}(f) = \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}$, where $P$ is the $x$ -component of $f$ and $Q$ is the $y$ -component. Let's differentiate! $\begin{aligned} \dfrac{\partial Q}{\partial x} &= \dfrac{\partial}{\partial x} \left[ y(\cos(x) + x) \right] \\ \\ &= -y\sin(x) + y \\ \\ \dfrac{\partial P}{\partial y} &= \dfrac{\partial}{\partial y} \left[ xy - 1 \right] \\ \\ &= x \end{aligned}$ Therefore: $\begin{aligned} \text{curl}(f) &= -y\sin(x) + y - x \end{aligned}$